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Kinematics question on a motorcycle

  1. Mar 4, 2007 #1
    1. The problem statement, all variables and given/known data

    A 100m long ramp is constructed perpendicular to a river’s edge with the
    highest point 20 m directly above the edge of the bank
    motorcycle stunt rider uses the ramp in an attempt to leap across a 30m wide
    river. The motorcycle accelerates from rest at the bottom of the ramp and takes
    off with a speed of 20 ms−1 moving parallel to the face of the ramp and
    perpendicular to the river bank.

    (a) What is the maximum height above the river banks reached by the
    motorcycle?
    (b) How long after take-off does the motorcycle land on the opposite bank?
    (c) Calculate how far from the river the motorcycle lands.
    (d) Calculate the velocity of the motorcycle at the moment of landing. Give your
    answer both as the magnitude and direction of the velocity vector and in
    component form.

    2. Relevant equations

    s= ut+1/2 at^2

    v2= u2 + 2as

    v= u+at



    3. The attempt at a solution

    (a)

    (a)

    arcsin0.2= 11.5369 degrees

    u= 20 sin 11.5369....= 3.9458...

    components in y diection-
    s=?
    v=0
    u=3.9458...
    a= -10

    v2= u2+2as
    o=15.569 + 2(-10) s
    20s= 15.569
    s= 0.77849 meteres.....which doesnt seem correct.....?

    can anyone do this ??
     
    Last edited: Mar 5, 2007
  2. jcsd
  3. Mar 4, 2007 #2

    hage567

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    Have you tried to calculate anything? You must show some work.
    Hint for a): What can you say about the velocity of the rider when he reaches maximum height?
     
  4. Mar 4, 2007 #3
    (a)
    at max height, v=0, v2= u2 + 2as
    0=400+ 2 (10) s
    -400=20s
    s=20 meters
    ??not sure
     
  5. Mar 4, 2007 #4

    hage567

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    No, that's not quite right. Since the rider is leaving the ramp at an angle, you must break up the components of his initial velocity into the x and y directions. You've been given enough information in the question to find that angle.
    Think about what is going on in each of these directions once he leaves the ramp. Write out the applicable kinematic equations for each direction so you can see how they fit together. v does equal zero at max height, but only for the y direction. Be careful with your signs, what is gravity doing to the rider as he reaches max height?
     
  6. Mar 4, 2007 #5
    height=20
    distance=30
    30/20= 1.5
    coz 1.5=?

    components in y diection-
    s=?
    v=0
    u=20
    a=10

    confused dont know what to do here..?
     
  7. Mar 4, 2007 #6

    hage567

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    The 30 m is the width of the river though, right? What is the length of the ramp? Draw out the triangle so you can see how to solve for the angle. Once you have that, you can apply it to the velocity to get the components (hint: draw another triangle). Then put your new value for u (in the y direction) into your equation.
     
  8. Mar 4, 2007 #7
    long ramp is constructed perpendicular to a river’s edge, this is kinda diagonal.


    length of ramp=100m
    inverse sin (20/100)=? give the angle= sin 0.2= ?

    components in y diection-
    s=?
    v=0
    u=20 inverse sin0.2
    a=10

    is this correct so far?
     
  9. Mar 4, 2007 #8

    hage567

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    "inverse sin (20/100)=? give the angle= sin 0.2= ?"

    This is backwards. If you take [tex]sin(theta)=20/100[/tex] from your triangle, then to find [tex]\theta[/tex] you would apply the inverse function to get [tex]\theta = arcsin(20/100)[/tex]. Do you understand that part? So knowing that, how is it going to change

    "u=20 inverse sin0.2"
    ?
     
  10. Mar 4, 2007 #9
    yes by mistake i wrote sin 0.2
    -----------------------------------
    arc sin 0.2 = theta i understand.
    this can be used for the velocity component.

    u= 20 arcsin 0.2 this will give vertical inital velocty.

    is that ok?
     
  11. Mar 4, 2007 #10
    which part of the quesiton have we done
    and
    are doing...?
     
  12. Mar 4, 2007 #11

    hage567

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    You found that [tex]\theta=\arcsin0.2[/tex]. What does this give you in degrees?

    OK, this is still not right, but try to work this out to get the actual magnitude for the velocity. Does the answer seem reasonable given what the total initial velocity is?
    I'm not sure if you don't understand how to find the components, or are just getting confused with notation. I will show you what u should be:
    By resolving the velocity vector you will find that for the y component, [tex]u=20\sin\theta[/tex]. So you have two options here. One is to just put in the number in degrees in for [tex]\theta[/tex] that I asked you to find above. The other is to put the expression for [tex]\theta[/tex] into this, getting [tex]u=20\sin(\arcsin0.2)[/tex]
    If you cannot see why the sin should be there, I would suggest reviewing vectors/trig from your text or notes.
    Work out the actual value for u. Does it seem reasonable?
    We haven't finished anything yet. Once you have this figured out, you can finish part (a) that you were working on before. This angle stuff is needed to do the question. This vector idea will also be needed again when you try to do part (d).

    See what you can come up with for the rest of the problem.
     
  13. Mar 5, 2007 #12
    (a)

    arcsin0.2= 11.5369 degrees

    u= 20 sin 11.5369....= 3.9458...

    components in y diection-
    s=?
    v=0
    u=3.9458...
    a= -10

    v2= u2+2as
    o=15.569 + 2(-10) s
    20s= 15.569
    s= 0.77849 meteres.....which doesnt seem correct.....?

    can anyone do this ??
     
  14. Mar 5, 2007 #13

    cristo

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    Staff Emeritus
    Science Advisor

    Why not? It seems correct to me. However, note that 20sin(theta) is exactly equal to 4 (since you worked out that sin(theta)=20/100)
     
  15. Mar 5, 2007 #14
    But 0.7784 meteres dosnt it seem to less for a bike to go up after a traveling at a fast speed...i would assume it should go about 2-3 meters?

    parts (b) (c) (d),,need help
     
    Last edited: Mar 5, 2007
  16. Mar 5, 2007 #15
    if u= 4
    then
    components in y diection-
    s=?
    v=0
    u=3.9458...
    a= -10

    v2= u2+2as
    o= 16 + 2(-10) s
    20s= 16
    s= 0.8 meteres

    (a) What is the maximum height above the river banks reached by the
    motorcycle? 0.8 meteres.
     
  17. Mar 5, 2007 #16

    hage567

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    The 0.8m is correct, but it is not the total height the motorcycle reached above the river. He jumped from a ramp that was above the ground, right?
     
  18. Mar 5, 2007 #17
    b.

    s= ut+1/2 at^2

    if u= 4
    then
    components in y diection-
    s=0.8
    u=4
    a= -10


    s= ut+1/2 at^2
    0.8 = 4t -5t^2

    8= 40t-50t^2

    4= 20t-25t^2

    -25t^2 + 20t - 4 = 0

    25t^2 - 20t + 4=0 solving quadratic equation

    20 + 200 / 50 = 4.4sec or

    20 - 200 / 50 = -3.6 secs

    as time cannot be minus therefore t= 4.4 secs
     
  19. Mar 5, 2007 #18
    total height = 20 + 0.8=
    20.8 meters...

    what about the rest of my answer part b.
     
  20. Mar 5, 2007 #19

    hage567

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    Why did you pick 0.8 m?? Is he 0.8 m off of the ground when he leaves the ramp? Think about the equation and about parabolic motion. You will also have to be careful with your signs in this (depending on which way you select positive/negative displacement).
     
  21. Mar 6, 2007 #20
    The 0.8m is correct, but it is not the total height the motorcycle reached above the river. He jumped from a ramp that was above the ground, right?

    thats the reason i chose 0.8 meters.

    im a bit confused now, can u help
     
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